1 2n. i=1. 2m+1 i = m, while. 2m + 1 = 1. is the best possible for any odd n.

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1 Problem 1. Let n be a positive integer. Find the largest nonnegative real number f(n) (depending on n) with the following property: whenever a 1, a,..., a n are real numbers such that a 1 + a + + a n is an integer, there exists some i such that a i 1 f(n). (Solution) The answer is { 0 if n is even, f(n) = if n is odd. 1 First, assume that n is even. If a i = 1 for all i, then the sum a 1 + a + + a n is an integer. Since a i 1 = 0 for all i, we may conclude f(n) = 0 for any even n. Now assume that n is odd. Suppose that a i 1 < 1 for all 1 i n. Then, since n a i is an integer, 1 n a i n n a i 1 < 1 n = 1, a contradiction. Thus a i 1 1 for some i, as required. On the other hand, putting n = m + 1 and a i = m for all i gives a m+1 i = m, while a i 1 = 1 m m + 1 = 1 (m + 1) = 1 for all i. Therefore, f(n) = 1 is the best possible for any odd n. Problem. Prove that every positive integer can be written as a finite sum of distinct integral powers of the golden mean τ = Here, an integral power of τ is of the form τ i, where i is an integer (not necessarily positive). (Solution) We will prove this statement by induction using the equality τ = τ + 1. If n = 1, then 1 = τ 0. Suppose that n 1 can be written as a finite sum of integral powers of τ, say k n 1 = a i τ i, (1) i= k where a i {0, 1} and n. We will write (1) as n 1 = a k a 1 a 0.a 1 a a k. () For example, 1 = 1.0 = 0.11 = =

2 Firstly, we will prove that we may assume that in () we have a i a i+1 = 0 for all i with k i k 1. Indeed, if we have several occurrences of 11, then we take the leftmost such occurrence. Since we may assume that it is preceded by a 0, we can replace 011 with 100 using the identity τ i+1 + τ i = τ i+. By doing so repeatedly, if necessary, we will eliminate all occurrences of two 1 s standing together. Now we have the representation n 1 = K i= K b i τ i, (3) where b i {0, 1} and b i b i+1 = 0. If b 0 = 0 in (3), then we just add 1 = τ 0 to both sides of (3) and we are done. Suppose now that there is 1 in the unit position of (3), that is b 0 = 1. If there are two 0 s to the right of it, i.e. n 1 = 1.00, then we can replace 1.00 with 0.11 because 1 = τ 1 + τ, and we are done because we obtain 0 in the unit position. Thus we may assume that n 1 = Again, if we have n 1 = , we may rewrite it as n 1 = = = and obtain 0 in the unit position. Therefore, we may assume that n 1 = Since the number of 1 s is finite, eventually we will obtain an occurrence of 100 at the end, i.e. n 1 = Then we can shift all 1 s to the right to obtain 0 in the unit position, i.e. and we are done. n 1 = , Problem 3. Let p 5 be a prime and let r be the number of ways of placing p checkers on a p p checkerboard so that not all checkers are in the same row (but they may all be in the same column). Show that r is divisible by p 5. Here, we assume that all the checkers are identical. ( ) p (Solution) Note that r = p. Hence, it suffices to show that p (p 1)(p ) (p (p 1)) (p 1)! 0 (mod p 4 ). (1)

3 Now, let f(x) := (x 1)(x ) (x (p 1)) = x p 1 + s p x p + + s 1 x + s 0. () Then the congruence equation (1) is same as f(p ) s 0 0 (mod p 4 ). Therefore, it suffices to show that s 1 p 0 (mod p 4 ) or s 1 0 (mod p ). Since a p 1 1 (mod p) for all 1 a p 1, we can factor x p 1 1 (x 1)(x ) (x (p 1)) (mod p). (3) Comparing the coefficients of the left hand side of (3) with those of the right hand side of (), we obtain p s i for all 1 i p and s 0 1 (mod p). On the other hand, plugging p for x in (), we get which implies f(p) = (p 1)! = s 0 = p p 1 + s p p p + + s 1 p + s 0, p p 1 + s p p p + + s p = s 1 p. Since p 5, p s and hence s 1 0 (mod p ) as desired. Problem 4. Let A, B be two distinct points on a given circle O and let P be the midpoint of the line segment AB. Let O 1 be the circle tangent to the line AB at P and tangent to the circle O. Let l be the tangent line, different from the line AB, to O 1 passing through A. Let C be the intersection point, different from A, of l and O. Let Q be the midpoint of the line segment BC and O be the circle tangent to the line BC at Q and tangent to the line segment AC. Prove that the circle O is tangent to the circle O. (Solution) Let S be the tangent point of the circles O and O 1 and let T be the intersection point, different from S, of the circle O and the line SP. Let X be the tangent point of l to O 1 and let M be the midpoint of the line segment XP. Since T BP = ASP, the triangle T BP is similar to the triangle ASP. Therefore, P T P B = P A P S. Since the line l is tangent to the circle O 1 at X, we have SP X = 90 XSP = 90 AP M = P AM which implies that the triangle P AM is similar to the triangle SP X. Consequently, XS XP = MP MA = XP MA and XP P S = MA AP.

4 From this and the above observation follows XS XP P T P B = XP MA P A P S = XP MA MA XP = 1. (1) Let A be the intersection point of the circle O and the perpendicular bisector of the chord BC such that A, A are on the same side of the line BC, and N be the intersection point of the lines A Q and CT. Since NCQ = T CB = T CA = T BA = T BP and CA Q = CAB = XAP = P AM = SP X, the triangle NCQ is similar to the triangle T BP and the triangle CA Q is similar to the triangle SP X. Therefore QN QC = P T P B and QC = XS QA XP. and hence QA = QN by (1). This implies that N is the midpoint of the line segment QA. Let the circle O touch the line segment AC at Y. Since ACN = ACT = BCT = QCN and CY = CQ, the triangles Y CN and QCN are congruent and hence NY AC and NY = NQ = NA. Therefore, N is the center of the circle O, which completes the proof. Remark : Analytic solutions are possible : For example, one can prove for a triangle ABC inscribed in a circle O that AB = k( + t), AC = k(1 + t), BC = k(1 + 4t) for some positive numbers k, t if and only if there exists a circle O 1 such that O 1 is tangent to the side AB at its midpoint, the side AC and the circle O. One obtains AB = k (1 + 4t ), AC = k (1 + t ), BC = k ( + t ) by substituting t = 1/4t and k = k t. So, there exists a circle O such that O is tangent to the side BC at its midpoint, the side AC and the circle O. In the above, t = tan 4R tan α α and k =, where R is the radius of O and A = α. Furthermore, t = tan γ and k = that tt = tan α tan γ = XS check that k = k t. XP P T (1+tan α)(1+4 tan α) 4R tan γ (1+tan γ)(1+4 tan γ), where C = γ. Observe = 1, which implies P B tt = 1. It is now routine easy to 4 Problem 5. In a circus, there are n clowns who dress and paint themselves up using a selection of 1 distinct colours. Each clown is required to use at least five different colours. One day, the ringmaster of the circus orders that no two clowns have exactly the same set

5 of colours and no more than 0 clowns may use any one particular colour. Find the largest number n of clowns so as to make the ringmaster s order possible. (Solution) Let C be the set of n clowns. Label the colours 1,, 3,..., 1. For each i = 1,,..., 1, let E i denote the set of clowns who use colour i. For each subset S of {1,,..., 1}, let E S be the set of clowns who use exactly those colours in S. Since S S implies E S E S =, we have E S = C = n, where S runs over all subsets of {1,,..., 1}. Now for each i, S E S E i if and only if i S, and hence E i = i S E S. By assumption, we know that E i 0 and that if E S, then S 5. From this we obtain ( ) E i = E S 5 E S = 5n. i S S Therefore n 48. Now, define a sequence {c i } 5 of colours in the following way: The first row lists c 1,..., c 1 in order, the second row lists c 13,..., c 4 in order, the third row lists c 5,..., c 36 in order, and finally the last row lists c 37,..., c 5 in order. For each j, 1 j 48, assign colours c j, c j+1, c j+, c j+3, c j+4 to the j-th clown. It is easy to check that this assignment satisfies all conditions given above. So, 48 is the largest for n. Remark : The fact that n 48 can be obtained in a much simpler observation that 5n 1 0 = 40. There are many other ways of constructing 48 distinct sets consisting of 5 colours. For example, consider the sets {1,, 3, 4, 5, 6}, {3, 4, 5, 6, 7, 8}, {5, 6, 7, 8, 9, 10}, {7, 8, 9, 10, 11, 1}, {9, 10, 11, 1, 1, }, {11, 1, 1,, 3, 4}, {1,, 5, 6, 9, 10}, {3, 4, 7, 8, 11, 1}. Each of the above 8 sets has 6 distinct subsets consisting of exactly 5 colours. It is easy to check that the 48 subsets obtained in this manner are all distinct.